Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
<em>Let </em><em>the </em><em>mass </em><em>be </em><em>X </em><em>g</em>
<em>percentage </em><em>=</em><em> </em><em>X/</em><em> </em><em>6.</em><em>5</em><em>0</em><em> </em><em>*</em><em> </em><em>100 </em><em>=</em><em>2.</em><em>2</em><em>%</em>
<em>X=</em><em> </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
<em>The </em><em>mass </em><em>is </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
Answer:
This reaction is characteristic to metal carbonates, which decompose when heated to form the oxide of the metal and carbon dioxide gas.
Explanation:
Just did it...
Your answer is going to be B. at the focal point of the lens. :)
